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Re:The Birthday Paradox – On Jupiter, and Beyond (2012) by resource0x on Sunday March 03, 2019 @07:43PM (#19297970)
Related: given a permutation of numbers 0..999, find out the position of a given number (e.g. number 5). It may seem that on average, you need 500 iterations. In fact, you need much less:

let index = 5; while (element[index] != 5) index = element[index];

Re:The Birthday Paradox – On Jupiter, and Beyond (2012) by pavel_lishin on Sunday March 03, 2019 @08:04PM (#19298045)
> In fact, you need much less

How many iterations does that take on average? And why?

Re:The Birthday Paradox – On Jupiter, and Beyond (2012) by resource0x on Sunday March 03, 2019 @08:57PM (#19298222)
see https://math.stackexchange.com/questions/1409862/average-len... Which gives you 1000/7.48, not 1000/2
Re:The Birthday Paradox – On Jupiter, and Beyond (2012) by saagarjha on Sunday March 03, 2019 @08:58PM (#19298230)
This seems to tend to 500 when I run this experimentally.
Re:The Birthday Paradox – On Jupiter, and Beyond (2012) by resource0x on Sunday March 03, 2019 @10:28PM (#19298558)
Indeed. Never tested it before, took the article in stackexchange for granted.